Answer:
1) Since you have not provided the equations to select the right one, I am going to explain you the relevant facts that are used to solve this question.
2) The transuranium elements are the chemiical elements with atomic number greater than that of the uranium.
The atomic number of uranium is 92. So, the transuranium elements are the elements with atomic number 93 or greater.
This are some of the transuranium elements:
Neptunio - 93
Plutonium - 94
Americium - 95
Curium - 96
Berkelium - 97
Californium - 98
Einstenium - 99
And so all the known elements (the last one is the 118).
3) In a nuclear reaction the total mass number ( shown as superscript to the left of the symbol) and total atomic number (shown as subscript to the left of the symbol) are conserved.
4) Beta decay is the release of a beta particle, which is an electron (considered massles and with charge - 1). So, the beta decay is represented with the symbol:
0
β, which means 0 mass and charge - 1.
-1
5) This is, then, an example of a β decay equation for one transuranium element:
239 239 0
Np → Pu + β
93 94 -1
As you see 239 = 239 + 0 and 93 = 94 - 1, showing that the total mass number ( shown as superscript to the left of the symbol) and the total atomic number (shown as subscript to the left of the symbol) are conserved.
Explanation:
Answer:
We can have: Calcium, strontium, or barium
Explanation:
In this case, we have to remember the solubility rules for sulfate
and the chloride
:
<u>Sulfate</u>
All sulfate salts are SOLUBLE-EXCEPT those also containing: Calcium, silver, mercury (I), strontium, barium or lead.(
), which are NOT soluble.
<u>Chloride</u>
All chloride salts as SOLUBLE-EXCEPT those also containing: lead, silver, or mercury (I). (
), which are NOT soluble.
If we the salt formed a precipitated with the sulfate anion, we will have as possibilities "Calcium, silver, mercury (I), strontium, barium or lead". If We dont have any precipitated with the Chloride anion we can discard "Silver, mercury (I), lead" and our possibilities are:
<u>"Calcium, strontium, or barium".</u>
I hope it helps!
The answer is: 1.5 moles of oxygen are present.
V(O₂) = 33.6 L; volume of oxygen.
p(O₂) = 1.0 atm; pressure of oxygen.
T = 0°C; temperature.
Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).
At STP one mole of gas occupies 22.4 liters of volume.
n(O₂) = V(O₂) ÷ Vm.
n(O₂) = 33.6 L ÷ 22.4 L/mol.
n(O₂) = 1.50 mol; amount of oxygen.
They can all by seperrated or replicated.